使用击键 A、 Ctrl + A、 Ctrl + C 和 Ctrl + V 的最大字符数

Here's how I would approach it:

• assume CtrlA = select all
• assume CtrlC = copy selection
• assume CtrlV = paste copied selection

given some text, it takes 4 keystrokes to duplicate it:

• CtrlA to select it all
• CtrlC to copy it
• CtrlV to paste (this will paste over the selection - STATE YOUR ASSUMPTIONS)
• CtrlV to paste again which doubles it.

From there, you can consider doing 4 or 5 A's, then looping through the above. Note that doing ctrl + a, c, v, v will grow your text exponentially as you loop through. If remaining strokes < 4, just keep doing a CtrlV

The key to interviews @ places like Google is to state your assumptions, and communicate your thinking. they want to know how you solve problems.

Using CtrlA + CtrlC + CtrlV is an advantage only after 4 'A's.

So I would do something like this (in pseudo-BASIC-code, since you haven't specified any proper language):

// We should not use the clipboard for the first four A's:
FOR I IN 1 TO MIN(N, 4)
PRINT 'CLICK A'
NEXT
LET N1 = N - 4


// Generates the maximum number of pastes allowed:
FOR I IN 1 TO (N1 DIV 3) DO
PRINT 'CTRL-A'
PRINT 'CTRL-C'
PRINT 'CTRL-V'
LET N1 = N1 - 3
NEXT


// If we still have same keystrokes left, let's use them with simple CTRL-Vs
FOR I IN N1 TO N
PRINT 'CTRL-V'
NEXT

Edit

1. Back to using a single CtrlV in the main loop.
2. Added some comments to explain what I'm trying to do here.
3. Fixed an issue with the "first four A's" block.

There's a dynamic programming solution. We start off knowing 0 keys can make us 0 A's. Then we iterate through for i up to n, doing two things: pressing A once and pressing select all + copy followed by paste j times (actually j-i-1 below; note the trick here: the contents are still in the clipboard, so we can paste it multiple times without copying each time). We only have to consider up to 4 consecutive pastes, since select, copy, paste x 5 is equivalent to select, copy, paste, select, copy, paste and the latter is better since it leaves us with more in the clipboard. Once we've reached n, we have the desired result.

The complexity might appear to be O(N), but since the numbers grow at an exponential rate it is actually O(N²) due to the complexity of multiplying the large numbers. Below is a Python implementation. It takes about 0.5 seconds to calculate for N=50,000.

def max_chars(n):
dp = [0] * (n+1)
for i in xrange(n):
dp[i+1] = max(dp[i+1], dp[i]+1) # press a
for j in xrange(i+3, min(i+7, n+1)):
dp[j] = max(dp[j], dp[i]*(j-i-1)) # press select all, copy, paste x (j-i-1)
return dp[n]

In the code, j represents the total number of keys pressed after our new sequence of keypresses. We already have i keypresses at this stage, and 2 new keypresses go to select-all and copy. Therefore we're hitting paste j-i-2 times. Since pasting adds to the existing sequence of dp[i] A's, we need to add 1 making it j-i-1. This explains the j-i-1 in the 2nd-last line.

Here are some results (n => number of A's):

• 7 => 9
• 8 => 12
• 9 => 16
• 10 => 20
• 100 => 1391569403904
• 1,000 => 3268160001953743683783272702066311903448533894049486008426303248121757146615064636953144900245 174442911064952028008546304
• 50,000 => a very large number!

I agree with @SB that you should always state your assumptions: Mine is that you don't need to paste twice to double the number of characters. This gets the answer for 7, so unless my solution is wrong the assumption must be right.

In case someone wonders why I'm not checking sequences of the form Ctrl+A, Ctrl+C, A, Ctrl+V: The end result will always be the same as A, Ctrl+A, Ctrl+C, Ctrl+V which I A4 consider.

Assume you have x characters in the clipboard and x characters in the text area; let's call it "state x".

Let's press "Paste" a few times (i denote it by m-1 for convenience), then "Select-all" and "Copy"; after this sequence, we get to "state m*x". Here, we wasted a total of m+1 keystrokes. So the asymptotic growth is (at least) something like f^n, where f = m^(1/(m+1)). I believe it's the maximum possible asymptotic growth, though i cannot prove it (yet).

Trying various values of m shows that the maximum for f is obtained for m=4.

Let's use the following algorithm:

Press A a few times
Press Select-all
Press Copy
Repeat a few times:
Press Paste
Press Paste
Press Paste
Press Select-all
Press Copy
While any keystrokes left:
Press Paste

(not sure it's the optimal one).

The number of times to press A at the beginning is 3: if you press it 4 times, you miss the opportunity to double the number of A's in 3 more keystrokes.

The number of times to press Paste at the end is no more than 5: if you have 6 or more keystrokes left, you can use Paste, Paste, Paste, Select-all, Copy, Paste instead.

So, we get the following algorithm:

If (less than 6 keystrokes - special case)
While (any keystrokes left)
A
Else
First 5 keystrokes: A, A, A, Select-all, Copy
While (more than 5 keystrokes left)
Paste, Paste, Paste, Select-all, Copy
While (any keystrokes left)
Paste

(not sure it's the optimal one). The number of characters after executing this is something like

3 * pow(4, floor((n - 6) / 5)) * (2 + (n - 1) % 5).

Sample values: 1,2,3,4,5,6,9,12,15,18,24,36,48,60,72,96,144,192,240,288,...

It takes 3 keystrokes to double your number of As. It only makes sense to start doubling when you have 3 or more As already printed. You want your last allowed keystroke to be a CtrlV to make sure you are doubling the biggest number you can, so in order to align it we will fill in any extra keystrokes after the first three As at the beginning with more As.

for (i = 3 + n%3; i>0 && n>0; n--, i--) {
print("a");
}


for (; n>0; n = n-3) {
print("ctrl-a");
print("ctrl-c");
print("ctrl-v");
}

Edit:

This is terrible, I completely got ahead of myself and didn't consider multiple pastes for each copy.

Edit 2:

I believe pasting 3 times is optimal, when you have enough keystrokes to do it. In 5 keystrokes you multiply your number of As by 4. This is better than multiplying by 3 using 4 keystrokes and better than multiplying by 5 using 6 keystrokes. I compared this by giving each method the same number of keystrokes, enough so they each would finish a cycle at the same time (60), letting the 3-multiplier do 15 cycles, the 4-multiplier do 12 cycles, and the 5-multiplier do 10 cycles. 3^15 = 14,348,907, 4^12=16,777,216, and 5^10=9,765,625. If there are only 4 keystrokes left, doing a 3-multiplier is better than pasting 4 more times, essentially making the previous 4 multiplier become an 8-multiplier. If there are only 3 keystrokes left, a 2-multiplier is best.

What follows uses the OP's second edit that pasting does not replace existing text.

Notice a few things:

• ^A and ^C can be considered a single action that takes two keystrokes, since it never makes sense to do them individually. In fact, we can replace all instances of ^A^C with ^K^V, where ^K is a one-key "cut" operation (let's abbreviate it X). We shall see that dealing with ^K is much nicer than the two-cost ^A^C.
• Let's assume that an 'A' starts in the clipboard. Then ^V (let's abbreviate it Y) is strictly superior to A and we can drop the latter from all consideration. (In the actual problem, if the clipboard starts empty, in what follows we'll just replace Y with A instead of ^V up until the first X.)

Every reasonable keystroke sequence can thus be interpreted as a group of Ys separated by Xs, for example YYYXYXYYXY. Denote by V(s) the number of 'A's produced by the sequence s. Then V(nXm) = V(n)*V(m), because X essentially replaces every Y in m with V(n) 'A's.

The copy-paste problem is thus isomorphic to the following problem: "using m+1 numbers which sum to N-m, maximimze their product." For example, when N=6, the answer is m=1 and the numbers (2,3). 6 = 2*3 = V(YYXYYY) = V(AA^A^C^V^V) (or V(YYYXYY) = V(AAA^A^C^V). )

We can make a few observations:

For a fixed value of m, the numbers to choose are ceil( (N-m)/(m+1) ) and floor( (N-m)/(m+1) ) (in whatever combination makes the sum work out; more specifically you will need (N-m) % (m+1) ceils and the rest floors). This is because, for a < b, (a+1)*(b-1) >= a*b.

Unfortunately I don't see an easy way to find the value of m. If this were my interview I would propose two solutions at this point:

Option 1. Loop over all possible m. An O(n log n) solution.

C++ code:

long long ipow(int a, int b)
{
long long val=1;
long long mul=a;


while(b>0)
{
if(b%2)
val *= mul;
mul *= mul;
b/=2;
}
return val;
}


long long trym(int N, int m)
{
int floor = (N-m)/(m+1);
int ceil = 1+floor;
int numceils = (N-m)%(m+1);
return ipow(floor, m+1-numceils) * ipow(ceil, numceils);
}


long long maxAs(int N)
{
long long maxval=0;
for(int m=0; m<N; m++)
{
maxval = std::max(maxval, trym(N,m));
}
return maxval;
}

Option 2. Allow m to attain non-integer values and find its optimal value by taking the derivative of [(N-m)/(m+1)]^m with respect to m and solving for its root. There is no analytic solution, but the root can be found using e.g. Newton's method. Then use the floor and ceiling of that root for the value of m, and choose whichever is best.

By using marcog's solution I found a pattern that starts at n=16. To illustrate this here are the keystrokes for n=24 up to n=29, I replaced ^A with S (select), ^C with C (copy), and ^V with P (paste) for readability:

24: A,A,A,A,S,C,P,P,P,S,C,P,P,P,S,C,P,P,P,S,C,P,P,P
4   *    4    *    4    *    4    *    4     = 1024
25: A,A,A,A,S,C,P,P,P,S,C,P,P,S,C,P,P,S,C,P,P,S,C,P,P
4   *    4    *   3   *   3   *   3   *   3    = 1296
26: A,A,A,A,S,C,P,P,P,S,C,P,P,P,S,C,P,P,S,C,P,P,S,C,P,P
4   *    4    *    4    *   3   *   3   *   3    = 1728
27: A,A,A,A,S,C,P,P,P,S,C,P,P,P,S,C,P,P,P,S,C,P,P,S,C,P,P
4   *    4    *    4    *    4    *   3   *   3    = 2304
28: A,A,A,A,S,C,P,P,P,S,C,P,P,P,S,C,P,P,P,S,C,P,P,P,S,C,P,P
4   *    4    *    4    *    4    *    4    *   3    = 3072
29: A,A,A,A,S,C,P,P,P,S,C,P,P,P,S,C,P,P,P,S,C,P,P,P,S,C,P,P,P
4   *    4    *    4    *    4    *    4    *    4     = 4096

After an initial 4 As, the ideal pattern is to select, copy, paste, paste, paste and repeat. This will multiply the number of As by 4 every 5 keystrokes. If this 5 keystroke pattern cannot consume the remaining keystrokes on its own some number of 4 keystroke patterns (SCPP) consume the final keystrokes, replacing SCPPP (or removing one of the pastes) as necessary. The 4 keystroke patterns multiply the total by 3 every 4 keystrokes.

Using this pattern here is some Python code that gets the same results as marcog's solution, but is O(1) edit: This is actually O(log n) due to exponentiation, thanks to IVlad for pointing that out.

def max_chars(n):
if n <= 15:
return (0, 1, 2, 3, 4, 5, 6, 9, 12, 16, 20, 27, 36, 48, 64, 81)[n]
e3 = (4 - n) % 5
e4 = n // 5 - e3
return 4 * (4 ** e4) * (3 ** e3)

Calculating e3: There are always between 0 and 4 SCPP patterns at the end of the keystroke list, for n % 5 == 4 there are 4, n % 5 == 1 there are 3, n % 5 == 2 there are 2, n % 5 == 3 there are 1, and n % 5 == 4 there are 0. This can be simplified to (4 - n) % 5.

Calculating e4: The total number of patterns increases by 1 whenever n % 5 == 0, as it turns out this number increases to exactly n / 5. Using floor division we can get the total number of patterns, the total number for e4 is the total number of patterns minus e3. For those unfamiliar with Python, // is the future-proof notation for floor division.

It's solveable in O(1): Like with the Fibonacci numbers, there is a formula to calculate the number of printed As (and the sequence of keystrokes):

1) We can simplify the problem description:

• Having only [A],[C-a]+[C-c],[C-v] and an empty copy-paste-buffer

equals

• having only [C-a]+[C-c],[C-v] and "A" in the copy-paste-buffer.

2) We can describe the sequence of keystrokes as a string of N chars out of {'*','V','v'}, where 'v' means [C-v] and '*' means [C-a] and 'V' means [C-c]. Example: "vvvv*Vvvvv*Vvvv"

The length of that string still equals N.

The product of the lengths of the Vv-words in that string equals the number of produced As.

3) Given a fixed length N for that string and a fixed number K of words, the outcome will be maximal iff all words have nearly equal lengths. Their pair-wise difference is not more than ±1.

Now, what is the optimal number K, if N is given?

4) Suppose, we want to increase the number of words by appending one single word of length L, then we have to reduce L+1 times any previous word by one 'v'. Example: "…*Vvvv*Vvvv*Vvvv*Vvvv" -> "…*Vvv*Vvv*Vvv*Vvv*Vvv"

Now, what is the optimal word length L?

(5*5*5*5*5) < (4*4*4*4*4)*4 , (4*4*4*4) > (3*3*3*3)*3

=> Optimal is L=4.

5) Suppose, we have a sufficient large N to generate a string with many words of length 4, but a few keystrokes are left; how should we use them?

• If there are 5 or more left: Append another word with length 4.

• If there are 0 left: Done.

• If there are 4 left: We could either

  a) append one word with length 3: 4*4*4*4*3=768.

  b) or increase 4 words to lenght 5: 5*5*5*5=625. => Appending one word is better.

• If there are 3 left: We could either

  a) or append one word with length 3 by adjusting the previus word from length 4 to 3: 4*4*4*2=128 < 4*4*3*3=144.

  b) increase 3 words to lenght 5: 5*5*5=125. => Appending one word is better.

• If there are 2 left: We could either

  a) or append one word with length 3 by adjusting the previus two words from length 4 to 3: 4*4*1=16 < 3*3*3=27.

  b) increase 2 words to lenght 5: 5*5=25. => Appending one word is better.

• If there is 1 left: We could either

  a) or append one word with length 3 by adjusting the previus three words from length 4 to 3: 4*4*4*0=0 < 3*3*3*3=81.

  b) increase one word to lenght 5: 4*4*5=80. => Appending one word is better.

6) Now, what if we don't have a "sufficient large N" to use the rules in 5)? We have to stick with plan b), if possible! The strings for small N are:

1:"v", 2:"vv", 3:"vvv", 4:"vvvv"

5:"vvvvv" → 5 (plan b)

6:"vvvvvv" → 6 (plan b)

7:"vvv*Vvv" → 9 (plan a)

8:"vvvv*Vvv" → 12 (plan a)

9:"vvvv*Vvvv" → 16

10:"vvvv*Vvvvv" → 20 (plan b)

11:"vvv*Vvv*Vvv" → 29 (plan a)

12:"vvvv*Vvv*Vvv" → 36 (plan a)

13:"vvvv*Vvvv*Vvv" → 48 (plan a)

14:"vvvv*Vvvv*Vvvv" → 64

15:"vvv*Vvv*Vvv*Vvv" → 81 (plan a)

…

7) Now, what is the optimal number K of words in a string of length N?

If N < 7 then K=1 else if 6 < N < 11 then K=2 ; otherwise: K=ceil((N+1)/5)

Written in C/C++/Java: int K = (N<7)?(1) : (N<11)?(2) : ((N+5)/5);

And if N > 10, then the number of words with length 3 will be: K*5-1-N. With this, we can calculate the number of printed As:

If N > 10, the number of As will be: 4^{N+1-4K}·3^{5K-N-1}

public int dp(int n)
{
int arr[] = new int[n];
for (int i = 0; i < n; i++)
arr[i] = i + 1;
for (int i = 2; i < n - 3; i++)
{
int numchars = arr[i] * 2;
int j = i + 3;
arr[j] = Math.max(arr[j], numchars);
while (j < n - 1)
{
numchars = numchars + arr[i];
arr[++j] = Math.max(arr[j], numchars);
}
}
return arr[n - 1];
}

Here is my approach and solution with code below.

Approach:

There are three distinct operations that can be performed.

1. Keystroke A - Outputs one character 'A'
2. Keystroke (Ctrl-A) + (Ctrl-C) - Outputs nothing essentially. These two keystrokes can be combined into one operation because each of these keystrokes individually make no sense. Also, this keystroke sets up the output for the next paste operation.
3. Keystroke (Ctrl-V) - Output for this keystroke really depends on the previous (second) operation and hence we would need to account for that in our code.

Now given the three distinct operations and their respective outputs, we have to run through all the permutations of these operations.

Assumption:

Now, some version of this problem states that the sequence of keystrokes, Ctrl+A -> Ctrl+C -> Ctrl+V, overwrite the highlighted selection. To factor in this assumption, only one line of code needs to be added to the solution below where the printed variable in case 2 is set to 0

        case 2:
//Ctrl-A and then Ctrl-C
if((count+2) < maxKeys)
{
pOutput = printed;


//comment the below statement to NOT factor
//in the assumption described above
printed = 0;
}

For this solution

The code below will print a couple of sequences and the last sequence is the correct answer for any given N. e.g. for N=11 this will be the correct sequence

With the assumption

A, A, A, A, A, C, S, V, V, V, V, :20:

Without the assumption

A, A, A, C, S, V, V, C, S, V, V, :27:

I have decided to retain the assumption for this solution.

Keystroke Legend:

'A' - A

'C' - Ctrl+A

'S' - Ctrl+C

'V' - Ctrl+V

Code:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>


void maxAprinted(int count, int maxKeys, int op, int printed, int pOutput, int *maxPrinted, char *seqArray)
{
if(count > maxKeys)
return;


if(count == maxKeys)
{
if((*maxPrinted) < printed)
{
//new sequence found which is an improvement over last sequence
(*maxPrinted) = printed;


printf("\n");
int i;
for(i=0; i<maxKeys; i++)
printf(" %c,",seqArray[i]);
}


return;
}


switch(op)
{
case 1:
//A keystroke
printed++;


seqArray[count] = 'A';
count++;
break;


case 2:
//Ctrl-A and then Ctrl-C
if((count+2) < maxKeys)
{
pOutput = printed;


//comment the below statement to NOT factor
//in the assumption described above
printed = 0;
}


seqArray[count] = 'C';
count++;
seqArray[count] = 'S';
count++;
break;


case 3:
//Ctrl-V
printed = printed + pOutput;


seqArray[count] = 'V';
count++;
break;
}


maxAprinted(count, maxKeys, 1, printed, pOutput, maxPrinted, seqArray);
maxAprinted(count, maxKeys, 2, printed, pOutput, maxPrinted, seqArray);
maxAprinted(count, maxKeys, 3, printed, pOutput, maxPrinted, seqArray);
}


int main()
{
const int keyStrokes = 11;


//this array stores the sequence of keystrokes
char *sequence;
sequence = (char*)malloc(sizeof(char)*(keyStrokes + 1));


//stores the max count for As printed for a sqeuence
//updated in the recursive call.
int printedAs = 0;


maxAprinted(0, keyStrokes,  1, 0, 0, &printedAs, sequence);


printf(" :%d:", printedAs);


return 0;
}

Using the tricks mentioned in answers above, Mathematically, Solution can be explained in one equation as,

4 + 4^[(N-4)/5] + ((N-4)%5)*4^[(N-4)/5]. where [] is greatest integer factor

There is a trade-off between printing m-A's manually, then using Ctrl+A, Ctrl+C, and N-m-2 Ctrl+V. The best solution is in the middle. If max key strokes = 10, the best solution is typing 5 A's or 4 A's.

try using this Look at this http://www.geeksforgeeks.org/how-to-print-maximum-number-of-a-using-given-four-keys/ and maybe optimize a bit looking for the results around the mid point.

Here is my solution with dynamic programming, without a nested loop, and which also prints the actual characters that you'd need to type:

N = 52


count = [0] * N
res = [[]] * N
clipboard = [0] * N


def maybe_update(i, new_count, new_res, new_clipboard):
if new_count > count[i] or (
new_count == count[i] and new_clipboard > clipboard[i]):
count[i] = new_count
res[i] = new_res
clipboard[i] = new_clipboard


for i in range(1, N):
# First option: type 'A'.
# Using list concatenation for 'res' to avoid O(n^2) string concatenation.
maybe_update(i, count[i - 1] + 1, res[i - 1] + ['A'], clipboard[i - 1])


# Second option: type 'CTRL+V'.
maybe_update(i, count[i - 1] + clipboard[i - 1],  res[i - 1] + ['v'],
clipboard[i - 1])


# Third option: type 'CTRL+A, CTRL+C, CTRL+V'.
# Assumption: CTRL+V always appends.
if i >= 3:
maybe_update(i, 2 * count[i - 3],  res[i - 3] + ['acv'], count[i - 3])


for i in range(N):
print '%2d %7d %6d %-52s' % (i, count[i], clipboard[i], ''.join(res[i]))

This is the output ('a' means 'CTRL+A', etc.)

 0       0      0
1       1      0 A
2       2      0 AA
3       3      0 AAA
4       4      0 AAAA
5       5      0 AAAAA
6       6      3 AAAacv
7       9      3 AAAacvv
8      12      3 AAAacvvv
9      15      3 AAAacvvvv
10      18      9 AAAacvvacv
11      27      9 AAAacvvacvv
12      36      9 AAAacvvacvvv
13      45      9 AAAacvvacvvvv
14      54     27 AAAacvvacvvacv
15      81     27 AAAacvvacvvacvv
16     108     27 AAAacvvacvvacvvv
17     135     27 AAAacvvacvvacvvvv
18     162     81 AAAacvvacvvacvvacv
19     243     81 AAAacvvacvvacvvacvv
20     324     81 AAAacvvacvvacvvacvvv
21     405     81 AAAacvvacvvacvvacvvvv
22     486    243 AAAacvvacvvacvvacvvacv
23     729    243 AAAacvvacvvacvvacvvacvv
24     972    243 AAAacvvacvvacvvacvvacvvv
25    1215    243 AAAacvvacvvacvvacvvacvvvv
26    1458    729 AAAacvvacvvacvvacvvacvvacv
27    2187    729 AAAacvvacvvacvvacvvacvvacvv
28    2916    729 AAAacvvacvvacvvacvvacvvacvvv
29    3645    729 AAAacvvacvvacvvacvvacvvacvvvv
30    4374   2187 AAAacvvacvvacvvacvvacvvacvvacv
31    6561   2187 AAAacvvacvvacvvacvvacvvacvvacvv
32    8748   2187 AAAacvvacvvacvvacvvacvvacvvacvvv
33   10935   2187 AAAacvvacvvacvvacvvacvvacvvacvvvv
34   13122   6561 AAAacvvacvvacvvacvvacvvacvvacvvacv
35   19683   6561 AAAacvvacvvacvvacvvacvvacvvacvvacvv
36   26244   6561 AAAacvvacvvacvvacvvacvvacvvacvvacvvv
37   32805   6561 AAAacvvacvvacvvacvvacvvacvvacvvacvvvv
38   39366  19683 AAAacvvacvvacvvacvvacvvacvvacvvacvvacv
39   59049  19683 AAAacvvacvvacvvacvvacvvacvvacvvacvvacvv
40   78732  19683 AAAacvvacvvacvvacvvacvvacvvacvvacvvacvvv
41   98415  19683 AAAacvvacvvacvvacvvacvvacvvacvvacvvacvvvv
42  118098  59049 AAAacvvacvvacvvacvvacvvacvvacvvacvvacvvacv
43  177147  59049 AAAacvvacvvacvvacvvacvvacvvacvvacvvacvvacvv
44  236196  59049 AAAacvvacvvacvvacvvacvvacvvacvvacvvacvvacvvv
45  295245  59049 AAAacvvacvvacvvacvvacvvacvvacvvacvvacvvacvvvv
46  354294 177147 AAAacvvacvvacvvacvvacvvacvvacvvacvvacvvacvvacv
47  531441 177147 AAAacvvacvvacvvacvvacvvacvvacvvacvvacvvacvvacvv
48  708588 177147 AAAacvvacvvacvvacvvacvvacvvacvvacvvacvvacvvacvvv
49  885735 177147 AAAacvvacvvacvvacvvacvvacvvacvvacvvacvvacvvacvvvv
50 1062882 531441 AAAacvvacvvacvvacvvacvvacvvacvvacvvacvvacvvacvvacv
51 1594323 531441 AAAacvvacvvacvvacvvacvvacvvacvvacvvacvvacvvacvvacvv

If N key Strokes are allowed, then the result is N-3.

A's -> N-3

CTRL+A -> Selecting those N Characters :+1

CTRL+C -> Copying those N Characters :+1

Ctrl+V -> Pasting the N Characters. :+1 i.e., (Since we have selected the whole characters using CTRL+A) Replacing these existing N-3 characters with the copied N-3 Characters(which is overriding the same characters) and the result is N-3.